The opts arguments may contain one or more of the following options.

disregard = <nothing (default), inequations, equations, all>. This option permits to restrict the test for the initialvalues (see above).

equations: the equations are disregarded; pdsolve only checks that the inequations are satisfied by initialvalues.

inequations: the inequations are disregarded; pdsolve only checks that the equations are satisfied by initialvalues.

all: equations and inequations are disregarded.

order = n. This option specifies the truncation order for the series expansion; if not given, the value of n used is that of Order.

The function call pdsolve(DE, series) returns an integral (exponents are nonnegative integers), formal (the convergence issue is not addressed), power series solution of the differential ideal defined by DE, truncated at order = n.

The unknowns may or not be specified. When the unknowns are specified (as a function, or a set or list of them), only these ones specified are considered unknowns and any other function is considered arbitrary. The solution is then expected to be valid for arbitrary values of these arbitrary functions. On the other hand, when the unknowns are not specified, all functions - differentiated or not - are considered unknowns.

The initial values for the unknowns, and/or their derivatives, and/or an expansion point, can be provided in DE by enclosing all the arguments into a set or a list. The set of initial values must have only one expansion point and must be linear in the unknowns and derivatives evaluated at the expansion point. Typically, these initial values are specified as equations of the form $u_0={\mathrm{value\_of\_u}}_0$, where the $u_0$ are dependent variables and/or their derivatives evaluated at the expansion point - say $u\left(0\&comma;t\right),u\left(x\&comma;0\right)$ or $u\left(0\&comma;0\right)$ when the unknown is $u\left(x\&comma;t\right)$ - and ${\mathrm{value\_of\_u}}_0$ represent their values, possibly being expressions involving independent variables (in which case keep the independent variable as is, in the left-hand-side $u_0$).

If initial values are not specified, the expansion point is the origin, all the independent variables equal to $0$, and all the functions and their derivatives different from zero. It is also possible to specify only an expansion point different than zero directly in DE - say including equations of the form $x=\mathrm{x0},t=\mathrm{t0}$.

The values attributed to the unknowns and their derivatives at the expansion point, also cannot be chosen freely. They must satisfy the system $F=0,S\ne 0$, returned by the DifferentialAlgebra command PowerSeriesSolution when using the option conditions. These conditions, which are checked by default, can be disregarded using the option disregard = ...; see the Examples section.

If the attributed values are solutions of the system $F=0,S\ne 0$, then, the returned power series are truncated solutions of the differential equation system DE.

If these values do not satisfy $F=0$, then, the returned power series are not truncated solutions of the differential ideal.

If they satisfy $F=0$, but do not satisfy $S\ne 0$, then, the returned power series may, or may not, be truncated solutions of the of DE. If they are, they do not need to be unique.

$\mathrm{pde}\left[1\right]≔\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;t\&comma;t\right)-\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;x\right)=0$

${\mathrm{pde}}_1≔\frac{{\partial }^2}{\partial t^2}u\left(x\&comma;t\right)-\frac{\partial }{\partial x}u\left(x\&comma;t\right)=0$

$\mathrm{iv}\left[1\right]≔u\left(x\&comma;0\right)=f\left(x\right),\mathrm{D}\left[2\right]\left(u\right)\left(x\&comma;0\right)=g\left(x\right)$

${\mathrm{iv}}_1≔u\left(x\&comma;0\right)=f\left(x\right),{\mathrm{D}}_2\left(u\right)\left(x\&comma;0\right)=g\left(x\right)$

$\mathrm{pdsolve}\left(\left[\mathrm{pde}\left[1\right]\&comma;\mathrm{iv}\left[1\right]\right]\&comma;\mathrm{series}\&comma;\mathrm{order}=2\right)$

$u\left(x\&comma;t\right)=\frac{\left(t^2+2x\right)\mathrm{D}\left(f\right)\left(0\right)}{2}+\mathrm{D}\left(g\right)\left(0\right)xt+\frac{{\mathrm{D}}^{\left(2\right)}\left(f\right)\left(0\right)x^2}{2}+g\left(0\right)t+f\left(0\right)$

$\mathrm{pde}\left[2\right]≔{\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;x\&comma;x\right)}^{2}v\left(x\&comma;t\right)+\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;x\&comma;x\right)v\left(x\&comma;t\right)+\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;x\right)=0,\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;x\&comma;t\right)=0,{\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;t\&comma;t\right)}^2-1=0$

${\mathrm{pde}}_2≔{\left(\frac{{\partial }^2}{\partial x^2}u\left(x\&comma;t\right)\right)}^{2}v\left(x\&comma;t\right)+\left(\frac{{\partial }^2}{\partial x^2}u\left(x\&comma;t\right)\right)v\left(x\&comma;t\right)+\frac{\partial }{\partial x}u\left(x\&comma;t\right)=0,\frac{{\partial }^2}{\partial t\partial x}u\left(x\&comma;t\right)=0,{\left(\frac{{\partial }^2}{\partial t^2}u\left(x\&comma;t\right)\right)}^2-1=0$

$\mathrm{iv}\left[2\right]≔v\left(x\&comma;0\right)=1+f\left(x\right),\mathrm{D}\left[2,2\right]\left(u\right)\left(0\&comma;0\right)=1,u\left(0\&comma;0\right)=c\left[0\right],\mathrm{D}\left[1\right]\left(u\right)\left(0\&comma;0\right)=0,\mathrm{D}\left[2\right]\left(u\right)\left(0\&comma;0\right)=c\left[2\right],\mathrm{D}\left[1,1\right]\left(u\right)\left(0\&comma;0\right)=-1$

${\mathrm{iv}}_2≔v\left(x\&comma;0\right)=1+f\left(x\right),{\mathrm{D}}_{2,2}\left(u\right)\left(0\&comma;0\right)=1,u\left(0\&comma;0\right)=c_0,{\mathrm{D}}_1\left(u\right)\left(0\&comma;0\right)=0,{\mathrm{D}}_2\left(u\right)\left(0\&comma;0\right)=c_2,{\mathrm{D}}_{1,1}\left(u\right)\left(0\&comma;0\right)=\mathrm{-1}$

$\mathrm{pdsolve}\left(\left[\mathrm{pde}\left[2\right]\&comma;\mathrm{iv}\left[2\right]\right]\&comma;\left[u\&comma;v\right]\left(x\&comma;t\right)\&comma;\mathrm{series}\&comma;\mathrm{order}=3\right)$

$\left\{u\left(x\&comma;t\right)=c_0+c_{2}t-\frac{1}{2}{t}^2\&comma;v\left(x\&comma;t\right)=1+f\left(0\right)+\mathrm{D}\left(f\right)\left(0\right)x+\frac{{\mathrm{D}}^{\left(2\right)}\left(f\right)\left(0\right)x^2}{2}+\frac{{\mathrm{D}}^{\left(3\right)}\left(f\right)\left(0\right)x^3}{6}\right\}$

$\mathrm{pde}\left[3\right]≔\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;x\&comma;x\right)=\exp \left(\sin \left(x\right)\right)\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;t\right)$

${\mathrm{pde}}_3≔\frac{{\partial }^2}{\partial x^2}u\left(x\&comma;t\right)={\&ExponentialE;}^{\sin \left(x\right)}\left(\frac{\partial }{\partial t}u\left(x\&comma;t\right)\right)$

$\mathrm{pdsolve}\left(\left\{\mathrm{pde}\left[3\right]\&comma;t=\mathrm{t0}\&comma;x=\mathrm{x0}\right\}\&comma;\mathrm{series}\&comma;\mathrm{order}=2\right)$

$u\left(x\&comma;t\right)=\frac{\left({\left(x-\mathrm{x0}\right)}^2{\&ExponentialE;}^{\sin \left(\mathrm{x0}\right)}+2t-2\mathrm{t0}\right){\mathrm{D}}_2\left(u\right)\left(\mathrm{x0}\&comma;\mathrm{t0}\right)}{2}+{\mathrm{D}}_{1,2}\left(u\right)\left(\mathrm{x0}\&comma;\mathrm{t0}\right)\left(x-\mathrm{x0}\right)\left(t-\mathrm{t0}\right)+\frac{{\mathrm{D}}_{2,2}\left(u\right)\left(\mathrm{x0}\&comma;\mathrm{t0}\right){\left(t-\mathrm{t0}\right)}^2}{2}+{\mathrm{D}}_1\left(u\right)\left(\mathrm{x0}\&comma;\mathrm{t0}\right)\left(x-\mathrm{x0}\right)+u\left(\mathrm{x0}\&comma;\mathrm{t0}\right)$

$\mathrm{pde}\left[4\right]≔\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;x\&comma;x\right)=\mathrm{BesselJ}\left(\mathrm{\nu }\&comma;\sin \left(x\right)\right)\mathrm{diff}\left(u\left(x\&comma;t\right)\&comma;t\right)$

${\mathrm{pde}}_4≔\frac{{\partial }^2}{\partial x^2}u\left(x\&comma;t\right)=\mathrm{BesselJ}\left(\mathrm{\nu }\&comma;\sin \left(x\right)\right)\left(\frac{\partial }{\partial t}u\left(x\&comma;t\right)\right)$

$\mathrm{pdsolve}\left(\left\{\mathrm{pde}\left[4\right]\&comma;x=\mathrm{x0}\right\}\&comma;\mathrm{series}\&comma;\mathrm{order}=2\right)$

$u\left(x\&comma;t\right)=\frac{\left({\left(x-\mathrm{x0}\right)}^2\mathrm{BesselJ}\left(\mathrm{\nu }\&comma;\sin \left(\mathrm{x0}\right)\right)+2t\right){\mathrm{D}}_2\left(u\right)\left(\mathrm{x0}\&comma;0\right)}{2}+{\mathrm{D}}_{1,2}\left(u\right)\left(\mathrm{x0}\&comma;0\right)\left(x-\mathrm{x0}\right)t+\frac{{\mathrm{D}}_{2,2}\left(u\right)\left(\mathrm{x0}\&comma;0\right)t^2}{2}+{\mathrm{D}}_1\left(u\right)\left(\mathrm{x0}\&comma;0\right)\left(x-\mathrm{x0}\right)+u\left(\mathrm{x0}\&comma;0\right)$